If a triangle has a height of 14 inches and a base of 9 inches, what's its area?
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Answer:
38.22% probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they score between 900 and 1,100, or they score outside this range. The probability of a student scoring between 900 and 1,100 is independent from other students. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
About 40 percent of these students’ scores were between 900 and 1,100.
This means that .
a)Based on this estimate, what is the probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100?
This is when
. So
In which
38.22% probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100
The equations are
,
The graphs of the solutions (x, y) of these equations are 2 parabolas, since the right hand side expressions are polynomials of degree 2.
The solution/s of the system are the x-coordinates of the point/s of intersection of the parabolas.
The solutions of the first equation form a parabola looking downwards (since the coefficient of x^2 is -), and the second, a parabola opening upwards (since the coefficient of x^2 is +).
We can draw both parabolas, but to find the solution we still need to solve the system algebraically.
The algebraic solution of the system is:
, so
the solutions are x=-1 and x=1.
The graph of the system is drawn using desmos.com
If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.
The graphs of the solutions (x, y) of these equations are 2 parabolas, since the right hand side expressions are polynomials of degree 2.
The solution/s of the system are the x-coordinates of the point/s of intersection of the parabolas.
The solutions of the first equation form a parabola looking downwards (since the coefficient of x^2 is -), and the second, a parabola opening upwards (since the coefficient of x^2 is +).
We can draw both parabolas, but to find the solution we still need to solve the system algebraically.
The algebraic solution of the system is:
the solutions are x=-1 and x=1.
The graph of the system is drawn using desmos.com
If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.